structural analysis

1. CEE-312
Structural Analysis and Design Sessional-I
(1.0 credit)
Lecture: 7&8
Bijit Kumar Banik
Assistant Professor, CEE, SUST
Room No.: 115 (“C” building)
bijit_sustbd@yahoo.com
Department of Civil and Environmental Engineering

2. Analysis and design of an Industrial roof truss system
Compr.TensionWL(R L)WL(L R)DL
-1.23+4.41-4.43-5.64+4.41L5L6
-1.23+4.41-4.43-5.64+4.41L4L5
Member
-0.66+3.53-4.19-3.11+3.53L3L4
Chord
-2.82+3.53-6.35-0.95+3.536.00L2L3
Bottom
-4.47+4.41-8.88-1.19+4.41L1L2
-4.47+4.41-8.88-1.19+4.41L0L1
-5.05+2.09+5.14+7.14-5.05L6U5
-4.04+1.57+5.00+5.61-4.04U4U5
Member
-3.03+1.82+4.85+4.07-3.03U3U4
Chord
-3.03+1.82+4.07+4.85-3.036.86U2U3
Top
-4.04+1.57+5.61+5.00-4.04U1U2
-5.05+2.09+7.14+5.14-5.05L0U1
Design Member Force (k)Member Force (k)Length
(ft)
MemberRemarks

3. Analysis and design of an Industrial roof truss system
-1.32+2.46+0.35+3.78-1.328.97L3U4
-0.93+2.14-3.07-3.07+2.1410.0L3U3
Member
-1.32+2.46+3.78+0.35-1.328.97L3U2
Web
-0.74+0.67-1.41-0.13+0.676.67L2U2
-1.01+1.89+2.90+0.27-1.016.86L2U1
-+0.1800+0.183.33L1U1
Compr.TensionWL(R L)WL(L R)DL
Length
(ft)
MemberRemarks
-+0.1800+0.183.33L5U5
-1.01+1.89+0.27+2.90-1.016.86L4U5
-0.74+0.67-0.13-1.41+0.676.67L4U4

4. Design bracing system
Bracing system consists of
a) Top chord bracing
b) Vertical bracing
c) Bottom chord strut
Exact analysis is seldom done in practice
Following guideline will be used
Tension member : KL/r < 400
Compression member: KL/r < 300
K = 0.7
Vertical & top chord bracing will be designed as tension member
Bottom chord strut will be designed as compression member

5. Industrial roof truss system
Vertical bracing
Bottom chord strut
Rise

6. Vertical bracing
Member of vertical bracing will be tied to each other at their crossing point.
So, half of length will be considered
Length of each member = 22
2010 + = 22.36 ft = 269 inch
400
)2/269(*7.0
<
r
r<
400
15.94
inchr 235.0>
From AISC chart, select
16
1
4
1
1
4
1
1 XXL ; for which rmin = 0.244

7. Design of Web chord members
1/8
0.291
0.984 in2
0.244

8. Industrial roof truss system
Span
Bay
Parlin
Top chord bracing
Sagrod
Column
Beam

9. Top chord bracing
Member of top chord bracing also tied to each other at their crossing point.
So, half of length will be considered
Length of each member = inchft 29125.2420)86.6*2( 22
==+
400
)2/291(*7.0
<
r
r<
400
85.101
inchr 255.0>
From AISC chart, select
8
1
2
1
1
2
1
1 XXL ; for which rmin = 0.298

10. Design of Web chord members
1/8
0.291
0.984 in2

11. Industrial roof truss system
Bottom chord strut
Rise
Tie

12. Bottom chord strut
To economize our design we will use lateral tie at the midspan of the strut, very
similar to sagrod. So, half of the bay length will be considered.
Length of strut ,
300
120*7.0
<
r
r<
300
0.84
inchr 28.0>
From AISC chart, select
8
1
2
1
1
2
1
1 XXL ; for which rmin = 0.298
inchL 12012*
2
20
==
Tie: ½ inch dia steel bar, connected with ½ inch nuts.

13. Design of Welded Connections
Here, All joints of the truss will be welded connections
Gusset plates will join the members at joints
Thickness of gusset plate will be assumed as the thickness of the
members
Two types of joints here-
a) Joints where all members end (L0, U3, L6)
b) Joints where there is one continuous member (L1, L2,…. L2 etc)
For all the joints we select the thickness of the gusset plate as 3/8 inch

14. Maximum forces on members
L0
L1 L2 L3 L4 L5
L6
U1
U2
U3
U4
U5
(- 4.47) (-4.47) (3.53) (3.53) (4.41) (4.41)
(-5.05)
(-4.04)
(-3.03) (-3.03)
(-4.04)
(-5.05)
(0.18)
(-0.74)
-0.74)
(0.18)
(2.14)
(2.46)
(1.89)
(2.46)
(1.89)
Type ‘a’ joints

15. Weld design of joint U3
16
3
22 XXL
8
3
2
1
1
2
1
1 XXL
16
3
22 XXL
U3
U 2
U 3 U
3 U
4
L3U3
P
=
3.03 k P = 3.03 k
P=2.14k
Gusset plate (3/8 in)
assumed

16. Gusset Plate

17. Weld design of joint U3
Allowable weld shear, FV = 0.3 Fy = 0.3 * 36 = 10.8 ksi
Weld design for member U2U3 or U3U4
P = 3.03 k
Weld size, s = 1/8 inch (assumed)
= 0.128 inch
Total weld length required, 71.3
125.0*707.0*8.10
03.3
707.0*
′′===
sF
P
L
v
So, L1+L2 = 3.17 inch

18. Weld design of joint U3
x
y
16
3
22 XXL
L1
L2
3.03 k

19. Weld design of joint U3
1/8
0.291
0.84 in2
3/16

20. Weld design of joint U3
0.569
1.431
16
3
22 XXL
L1
L2
3.03 k
515.2
569.0
431.1
2
1
==
L
L
L1 = 2.515 L2
L1+L2 = 3.17 inch
2.515 L2 + L2 = 3.17
L2 = 0.902 ≈ 1 inch
L1 = 0.902* 2.515 = 2.27 ≈ 2.5 inch
Same way calculate U3L3

21. Weld design of joint U3
16
3
22 XXL
8
3
2
1
1
2
1
1 XXL
16
3
22 XXL
U3
U 2
U 3 U
3 U
4
L3U3
P
=
3.03 k P = 3.03 k
P=2.14k
Gusset plate (3/8 in)
assumed
1/8 1
1/82.5

22. Weld design of joint L3
For L2L3L4 continuous bottom chord member
2.16(+3.53-3.11)
= -0.42
(+3.53-0.95)
= -2.58
DL+WL
(L→R)
2.16(+3.53-4.19)
= -0.66
(+3.53-6.35)
= -2.82
DL+WL
(R→L)
0+3.53+3.53Dead load
Resultant
[A~B]
L3L4(member
force)
B
L2L3(member
force)
A
Condition
P = 2.16 k Now, same as previous !!